In: Statistics and Probability
A study looked at if TV show content influenced the ability of watchers to remember brand names of items in commercials. Researchers randomly assigned volunteers to watch one of three programs, each containing the same nine commercials. After the show ended, the volunteers were asked to remember the brands of products that were advertised. Below shows how many of the nine commercials the volunteers remembered correctly:
Program | Size | Mean | SD
Violent | 108 | 2.08 | 1.87
Sexual | 108 | 1.71 | 1.76
Neutral | 108 | 3.17 | 1.77
Perform an hypothesis test. State the hypotheses, t-stat, p-value, and conclusion.
Using a Bonferroni adjustment, find simultaneous 95% confidence intervals for the pairwise differences in means, and refine your conclusion from part (1). (so a set of 3 confidence intervals.)
One Way Analysis of Variance (ANOVA)
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | std dev | |
A | 108 | 224.64 | 2.080 | 3.497 | 1.8700 | |
B | 108 | 184.68 | 1.710 | 3.098 | 1.7600 | |
C | 108 | 342.36 | 3.170 | 3.133 | 1.7700 | |
#VALUE! | ||||||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 124.438 | 2 | 62.219 | 19.189 | 0.0000 | 3.02 |
Within Groups | 1040.832 | 321 | 3.242 | |||
Total | 1165.269 | 323 |
Bonferroni adjustment
α=0.05/3 = 0.0167
Level of significance= | 0.0167 |
no. of treatments,k= | 3 |
DF error =N-k= | 321 |
MSE= | 3.2425 |
t-critical value,t(α/2,df)= | 2.4066 |
confidence interval | ||||||||
population mean difference | critical value | lower limit | upper limit | result | test stat | p value | ||
µ1-µ2 | 0.370 | 0.58972 | -0.2197 | 0.9597 | means are not different | 1.510 | 0.1320 | |
µ1-µ3 | -1.090 | 0.59 | -1.68 | -0.50 | means are different | -4.448 | 0.0000 | |
µ2-µ3 | -1.460 | 0.59 | -2.05 | -0.87 | means are different | -5.958 | 0.0000 |